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Because $S[x_0]=0$ at $x=0$, the factor $Z$ cancels from the ratio. Write $z=\beta \hslash \omega$ and $q=\sqrt{m\omega /\hslash }x$. The exact answer isAt high temperature, $z\ll 1$. Expanding the hyperbolic tangent givesThe displayed leading expression is the ratio of classical Boltzmann factors for potential energy $V(x)=m{\omega }^2{x}^2/2$ and $V(0)=0$. Indeed the classical canonical ensemble givesThe momentum Gaussian integral is independent of $x$ and cancels in the ratio. This is the classical phase-space limit of the thermal kernel. Here $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant; the associated classical position variance is $1/(\beta m{\omega }^2)$. At fixed $x$, $R$ ultimately tends to one as temperature tends to infinity, while its width grows.

At low temperature, $z\gg 1$. Now $\tanh (z/2)=1-2e^{-z}+O(e^{-2z})$, so at fixed $q$Unlike the classical high-temperature expression, this limiting width remains nonzero at zero temperature, reflecting the ground state of the quantum harmonic oscillator.

Insert the complete orthonormal basis of energy eigenstates on either side of the Boltzmann factor. The spectral theorem givesAt low temperature, the gap $\hslash \omega$ suppresses excited energy eigenstates. The harmonic oscillator eigenfunction of the ground state isThereforeThe common zero-point energy factor cancels. Dividing the kernel by its trace would give the position probability density of the canonical ensemble; at zero temperature this tends to $|{\phi }_0(x){|}^2$, whose variance is $\hslash /(2m\omega )$. Thus zero temperature projects onto the ground state and retains its quantum spread, rather than concentrating at the minimum of the classical potential energy.

There is also a check on the first correction. With $q=\sqrt{m\omega /\hslash }x$, the first excited harmonic oscillator eigenfunction satisfies $|{\phi }_1(x){|}^2=2q^2|{\phi }_0(x){|}^2$ and ${\phi }_1(0)=0$. Hence the spectral theorem gives $R=e^{-q^2}[1+2q^2{e}^{-z}+O(e^{-2z})]$, in agreement with part (c).

Each neighboring pair of Ising spins contributes at least $-J$ to the energy, with equality exactly when the two Ising spins are opposite. For even $N$, alternation closes consistently under the periodic boundary conditions. The ground states are thereforeHere $g_0$ is the ground-state degeneracy. Choosing $S_1$ fixes every other Ising spin, so these are all the ground states; translations do not give additional spin configurations. The two ground states are exchanged by spin inversion symmetry.

Keep $N$ fixed and put $q=\tanh K$, so $q\to 1^{-}$ as temperature tends to zero. For even $N$, the finite-ring antiferromagnetic Ising correlation becomesBoth alternating ground states have $S_1{S}_{n+1}=(-1{)}^n$, so their equal-weight canonical ensemble has the same two-point correlation function despite its zero one-point correlation function.

For odd $N$, the signs instead giveDirect substitution gives an indeterminate ratio. Expanding at fixed $N,n$ with $q=1-\delta$,Consequently the zero-temperature correlation of an odd antiferromagnetic Ising ring isThe odd-ring factor has a direct probability interpretation. At zero temperature, the canonical ensemble gives equal weight to its $2N$ ground states. The single frustrated bond in an antiferromagnetic Ising ring is uniformly located, with probability $n/N$ of lying on the chosen $n$-bond path. If it is absent from that path, $S_1{S}_{n+1}=(-1{)}^n$; if it is present, this product is $-(-1{)}^n$. Averaging gives $(-1{)}^n[(1-n/N)-n/N]$.

Thus the decay with separation here comes from averaging a movable, compulsory frustrated bond in an antiferromagnetic Ising ring, even after additional thermally excited frustrated bonds in an antiferromagnetic Ising ring disappear. In particular $C_1=-1+2/N$ for odd $N$, and $JN{C}_1=-J(N-2)$ reproduces the ground-state energy. The formula also respects $C_n=C_{N-n}$, although its factor $1-2n/N$ changes sign halfway around the ring.

The order and scale of limits matter. Taking the thermodynamic limit first at positive temperature gives the exponential function in part (c). Cooling an odd finite ring first leaves the linear factor $1-2n/N$. Both orders subsequently give $(-1{)}^n$ at fixed $n$, but this agreement is not uniform in separation: when $n/N$ stays nonzero, the odd-ring linear factor survives in the zero-temperature expression.

The canonical anticommutation relations make $(a^{†}{b}^{†}{)}^2=0$, so the proposed exponential function terminates exactly:Use the inverse Bogoliubov transformation $c=ua-v{b}^{†}$, $d=v{a}^{†}+ub$, and the actions $a|11⟩=b^{†}|0⟩$ and $b|11⟩=-a^{†}|0⟩$. The two vacuum state conditions becomeBoth therefore require $B=v/u$. Normalizing this quantum state gives $1=A^2(1+B^2)$. Choosing positive $A$, which fixes an irrelevant overall global phase, yields $A=u$ because $u^2+v^2=1$. Thus the two-mode fermionic paired vacuum isThe other sign for $A$ describes the same physical quantum state. The positive relative coefficient is fixed by the paper's pairing order and the convention $|11⟩=a^{†}{b}^{†}|0⟩$.

As a direct eigenvector check, $\Delta v=(E-\epsilon )u$ and $\Delta u=(E+\epsilon )v$ implyIt is therefore the unique ground state, as also follows from positive quasiparticle excitation energy $E$. It has definite even fermion parity but is not an eigenstate of the original total number operator: the empty-pair and occupied-pair probabilities are $u^2$ and $v^2$. In particular $⟨N_a⟩=⟨N_b⟩=v^2$. The pair amplitude is $⟨ab⟩=-uv$, consistently with $ab|11⟩=-|0⟩$.

For the limiting check $\Delta \to 0^{+}$ at fixed positive $\epsilon$, $A\to 1$, $B\to 0$, so $|\tilde{0}⟩\to |0⟩$. For $\epsilon \to 0^{+}$ at fixed positive $\Delta$, $A\to 1/\sqrt{2}$, $B\to 1$, giving equal amplitudes for the empty and occupied pair. These checks also fix which branch of the Bogoliubov transformation produces the ground state.

Cut one nonseparating circle on each handle of the closed orientable surface ${\Sigma }_g$ of genus $g$. The cut surface is a sphere with $2g$ boundary circles. Gluing a pair back together requires an anyon label $a_i$ on one boundary and the antianyon label ${\bar{a}}_i$ on the oppositely oriented boundary. Sum over these labels independently for the $g$ handles.

For a fixed labeling, the dimension of the resulting anyon fusion space on the sphere, with total vacuum anyon charge, isThe gluing rule of the topological quantum field theory therefore gives the handle-cut formula for anyon ground-state degeneracy:Here $1$ denotes the vacuum anyon, and the anyon fusion matrices have entries $(N_b{)}_{ac}=N_{ab}^c$, as in the source. At $g=0$, the empty product is the identity matrix, giving one ground state. At $g=1$, $(N_a{N}_{\bar{a}}{)}_{11}=1$ for each simple charge, so the ground-state degeneracy of the torus is the number of anyon types.

A hydrogen recombination line is a discrete spectral line produced when radiative recombination captures an electron into an excited bound level and a subsequent radiative cascade emits photons. The free-to-bound capture itself produces a continuum; the later bound-to-bound transitions produce the recombination lines. For example, H-alpha is the $n=3\to 2$ transition in the Balmer series.

For the neutral hydrogen absorbing an ionizing stellar continuum, the principal bound-free transition is photoionization from the $1s$ ground state into the continuum. Most neutral hydrogen is in its ground state, and the photoionization cross section is largest near its $13.6\mathrm{eV}$ threshold, falling approximately as ${\nu }^{-3}$ above it. Dominance of the population-weighted opacity is intended here; this does not mean that every excited-state microscopic cross section is smaller at every frequency.

In an ionization-bounded, optically thick nebula, direct radiative recombination into $1s$ emits an ionizing photon that promptly photoionizes another hydrogen atom. One ion disappears and another is created: there is no net removal of an ion from the global budget. Case B recombination therefore uses ${\alpha }_B={\sum }_{n\ge 2}{\alpha }_n$, with local recycling of ground-state recombination photons. This is the on-the-spot approximation, rather than an unconditional property of radiative recombination: if those photons escape, direct capture does produce net neutralization and a Case A treatment is appropriate.

Use the condensate wavefunction convention $\int |\psi {|}^2{d}^{3}r=N$, so $n(\mathbf{r})=|\psi (\mathbf{r}){|}^2$ is the number density and $\psi /\sqrt{N}$ is a normalized one-particle orbital. The contact interaction strength is $U_0=4\pi {\hslash }^{2}a/m$ for dilute atoms with s-wave scattering length $a$. The chemical potential $\mu =\partial E/\partial N$ fixes the stationary time dependence $e^{-i\mu t/\hslash }$ and enforces normalization when minimizing the energy. The Gross–Pitaevskii equation follows from an energy functional that contains the pair-counting factor $1/2$:Varying $E-\mu \int |\psi {|}^2{d}^{3}r$ gives the stationary Gross–Pitaevskii equation; the interaction contribution to that equation is $U_{0}n\psi$, without the factor $1/2$.

For the broadening discussed here assume repulsive interactions, $a>0$. The Thomas–Fermi approximation for a condensate neglects the gradient kinetic energy compared with the trapping and interaction energies. At fixed spatial scale, increasing $N{U}_0$ increases the mean-field potential. Expansion then reduces the kinetic energy further: for a cloud radius $R$, its scale per atom is ${\hslash }^2/(m{R}^2)$, whereas the trap energy scale is $m{\omega }^2{R}^2$. Their ratio decreases as $(a_0/R{)}^4$. In the weak-interaction limit the kinetic energy instead balances confinement, giving the Gaussian wave packet ground state of the quantum harmonic oscillator.

The spherical Thomas–Fermi condensate hasIntegrating this density gives $N=8\pi \mu R^3/(15U_0)$. ThusAt fixed $N$, stronger repulsion produces a wider, lower wavefunction: the Gaussian wave packet crosses over towards the square root of an inverted parabola, with an almost flat central region and a steeper outer drop. The density is the inverted parabola; the wavefunction is its square root. The exact wavefunction remains smooth and has a tail, since kinetic energy cannot be neglected near the Thomas–Fermi condensate edge rounding region. The healing length must be small compared with the cloud size in the bulk. Attractive interactions would not follow this large-$N$ expansion argument.

The original plot compares the noninteracting endpoint with two normalized repulsive spherical Thomas–Fermi condensate profiles. The latter are asymptotic shapes, rather than exact intermediate solutions of the Gross–Pitaevskii equation. Each has the same integrated atom number; $\chi =Na/a_0$ labels the illustrative interaction strengths.

Choose $A$ real and positive. The Gaussian integral givesThe three zero-point energies add for each atom. The ideal-gas chemical potential is a one-atom energy; the total energy includes all $N$ atoms.

For weak interactions, the first-order correction in perturbation theory is the energy functional of the contact interaction evaluated on the unperturbed Gaussian wave packet. The integral needed isConsequently the Gaussian interaction energy of a trapped condensate isHere $\hslash \omega ={\hslash }^2/(m{a}_0^2)$. These are the mean-field expressions consistent with the supplied equation and $\int |\psi {|}^2=N$. Exact first-order perturbation theory for precisely $N$ atoms in one orbital replaces $N^2$ by $N(N-1)$ through the number of distinct pairs; this distinction is negligible in the large-$N$ regime of the crossover estimate.

To judge the ideal-gas approximation, compare the interaction energy with the total ideal energy, or compare the interaction energy per atom with $\hslash \omega$:Thus the crossover from Gaussian to Thomas–Fermi condensate is governed by $\chi =Na/a_0$: perturbation theory is reliable for $\chi \ll 1$, while the regime of the Thomas–Fermi approximation for a condensate requires $\chi \gg 1$. The characteristic crossover number is $N_c\sim a_0/a$; there is no sharp critical atom number here. If one specifically defines crossover by $E_I=E_0$ using the weak-interaction expression, it gives $N_c=(3\sqrt{2\pi }/2)a_0/a$. That prefactor is a convention-dependent estimate extrapolated to the edge of perturbation theory, not a phase transition. Comparing $E_I$ itself to a single $\hslash \omega$ would incorrectly omit a factor of $N$.

For the numerical estimate use $m\simeq 87\times 1.66054\times 10^{-27}\mathrm{kg}$ and $\hslash \simeq 1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$. ThenThe particular equality $E_I=E_0$ would instead give about $1.15\times 10^3$ atoms. These quantify different choices within a smooth crossover; a well-developed spherical Thomas–Fermi condensate profile requires going substantially beyond the characteristic scale.

Fix the qubit polarization convention by $\rho =(I+\vec{P}\cdot \vec{\sigma })/2$, where the Pauli matrices are ordered in the basis $(|1⟩,|2⟩)$ and ${\sigma }_z=|1⟩⟨1|-|2⟩⟨2|$. Thus the ground state is the north pole of the Bloch sphere. In a rotating frame and within the rotating-wave approximation, choose the drive phase and detuning sign so thatFor example, with this ground-state convention one may take $\Delta ={\omega }_L-{\omega }_0$ and choose the transverse coupling positive. The Bloch vector precesses around the effective field $(\Omega ,0,\Delta )$ at angular frequency ${\Omega }_R=\sqrt{{\Omega }^2+{\Delta }^2}$, tracing a circle at fixed projection on that axis. Its length is conserved under coherent unitary time evolution. On resonance the axis is transverse and the Rabi oscillation can fully invert an initially ground-state atom. Away from resonance the axis tilts, andThere is no relaxation towards an equilibrium Bloch vector in this ideal model. An inhomogeneous ensemble can nevertheless lose its averaged qubit polarization through dephasing even without spontaneous emission; the circle description assumes a common drive and detuning.

For $|a{|}^2+|b{|}^2=1$, the stated quantum state givesThe diagonal elements are the two state populations. The off-diagonal elements are quantum coherences: they retain the relative quantum phase and determine interference under a subsequent pulse. This is a pure state, even if both populations are nonzero: ${\rho }^2=\rho$, $\mathrm{Tr}{\rho }^2=1$, and $|\vec{P}|=1$. A mixed state averages projectors over preparation probabilities and generally has $\mathrm{Tr}{\rho }^2<1$. Equal populations alone do not distinguish an equal coherent superposition from an unpolarized mixture.

For use in all three cases, multiplication of the gates gives the Hadamard-phase-Hadamard qubit rotationThe global phase cancels in ${\rho }^{\prime }=U\rho U^{†}$. Thus the Bloch vector map is

The final unlabelled Ramsey spectroscopy comparison applies to the three cases as follows. In an atomic clock, an initial coherent pulse prepares a superposition, a free interval $T$ accumulates a relative quantum phase proportional to detuning times $T$, and a final coherent pulse converts that phase to a population. Here $\alpha$ plays the role of that accumulated phase, with sign and fringe offset set by pulse conventions. In (a), the first Hadamard gate creates the needed superposition and the final gate reveals full-contrast Ramsey interferometry. In (b), the first Hadamard gate instead puts the state in $|1⟩$, so the intervening phase gate has no relative quantum phase to imprint: the output is independent of $\alpha$. In (c), the unpolarized density matrix remains unchanged under every gate, so the ensemble has no Ramsey interferometry contrast. A coherent unitary gate alone cannot initialize a maximally mixed ensemble into a pure clock state.

The Ramsey interferometry analogy concerns preparation, phase accumulation, and readout; a Hadamard gate is not literally a transverse $\pi /2$ rotation on every input. In this basis $H=R_y(\pi /2){\sigma }_z$, and its Bloch sphere action is a $\pi$ rotation about $(\hat{x}+\hat{z})/\sqrt{2}$, up to a global phase. The phase conventions of physical Ramsey spectroscopy pulses must be specified to compare their outputs for arbitrary initial states.

For a free-space electric dipole transition, the spontaneous emission rate in SI units isHere $|\mathbf{d}{|}^2$ includes the relevant dipole components; a specified single polarization requires its corresponding matrix element.

A complete scattering cycle returns the internal state to the ground state and changes the atomic momentum to $\mathbf{P}+\mathbf{p}-{\mathbf{p}}^{\prime }$. ThusFor photon-scattering recoil heating with isotropic emission of fixed momentum magnitude, $⟨{\mathbf{p}}^{\prime }⟩=0$, givingThe first term is the velocity-dependent work of absorption and can cool an atom if the photon opposes its motion. The last two are positive absorption and emission recoil energies. If the actual absorbed directions are also isotropic and independent of $\mathbf{P}$, the first term averages to zero. For approximately equal incoming and outgoing photon momenta $\hslash k$,Isotropic incident illumination alone would not guarantee isotropic absorption if Doppler-dependent excitation selected directions.

Source qualification for the trapped calculation. Set $m=M$ for the same atom. The printed Gaussian $\exp [-x^2/(2l_0{)}^2]$ has the wrong width for the specified harmonic oscillator: the printed ground and first-excited functions have squared norms $\sqrt{2}$ and $2\sqrt{2}$, respectively. The normalized oscillator states fixed by the given $l_0=\sqrt{\hslash /(m{\omega }_{\mathrm{HO}})}$ instead useAlso, one common complex phase multiplying both off-diagonal terms does not make the Hamiltonian a Hermitian operator. Use an absorption matrix element $(\hslash \Omega /2)e^{-ikx}$ and its Hermitian conjugate for emission; choosing the opposite propagation convention reverses the imaginary sideband phase but not a transition probability. The physical results below use these explicitly stated corrections. Parts (i) and (ii) also give the literal integrals of the printed functions.

Conditional motional heating and recoil averaging. With $\eta =k{l}_0/\sqrt{2}$, an oscillator eigenstate has $⟨K{⟩}_n=⟨V{⟩}_n=(n+1/2)\hslash {\omega }_{\mathrm{HO}}/2$. Therefore the absorption outcomes haveThe same statements hold for the specified emission outcomes $|e,0⟩\to |g,0⟩$ and $|e,0⟩\to |g,1⟩$. Internal excitation energy goes into, or comes from, the photon and is distinct from this external energy change. Emission into the first motional state makes the emitted photon correspondingly less energetic.

A sudden recoil from $|0⟩$ gives probabilities $P_n=e^{-{\eta }^2}{\eta }^{2n}/n!$, so $P_1\simeq {\eta }^2\ll 1$ while $P_0\simeq 1$. The rare first sideband transfers a whole trap quantum, much larger than $E_R$; the mean is neverthelessThe emission average is the same for a one-dimensional recoil of the same magnitude. Projecting isotropic three-dimensional emission onto one axis instead gives a factor $⟨{\cos }^2\theta ⟩=1/3$ for that axis.

A sudden unobserved kick initially raises kinetic energy by $E_R$ and leaves the position distribution, hence potential energy, unchanged. It creates a coherent superposition of motional states. The equal kinetic/potential increments above concern energy-eigenstate-conditioned outcomes, or time-averaged/dephased motion; these are different measurement conditions. There is no contradiction.

With $\Gamma >{\omega }_{\mathrm{HO}}$, motional sidebands are not cleanly resolved, and a short-lived excited atom has little time to move before decay. In the stronger sudden limit $\Gamma \gg {\omega }_{\mathrm{HO}}$ the recoil-kick picture is direct. For only the stated inequality, finite linewidth weights can change numerical factors; it does not fix an exact continuous-wave scattering rate. The average heating remains of recoil order per absorption or emission, and of order $2E_R$ when both absorption and emission kicks are counted at full magnitude. For the axial component of isotropic three-dimensional emission, the emission contribution instead carries the factor $1/3$ noted above. Strong confinement suppresses the chance of changing $n$, but does not make its averaged recoil energy disappear. This agrees with free-space recoil heating while replacing small continuous energy changes by rare quantized motional changes.

A Jablonski diagram distinguishes electronic excitation from subsequent relaxation. Photon absorption takes a molecule from the electronic ground state $S_0$ to a vibrationally excited level of $S_1$ or a higher singlet state. Rapid vibrational relaxation and internal conversion usually populate the lowest vibrational levels of $S_1$. Fluorescence then emits a photon on a transition to $S_0$, often ending in one of its excited vibrational levels. Nonradiative relaxation competes with emission; intersystem crossing can also populate triplet states. Energy lost to relaxation and solvent reorganization commonly makes the fluorescence maximum occur at lower photon energy, hence longer wavelength, than the photon absorption maximum: the Stokes shift.

The intrinsic fluorescence quantum yield is the number of photons emitted by fluorescence per absorbed excitation. For a simple excited state with radiative rate $k_r$ and other decay rate $k_{nr}$,A nearby acceptor adds a nonradiative energy-transfer channel, not an emitted photon flying from donor to acceptor. The electric dipole-dipole interaction transfers the donor excitation while returning the donor toward its ground state and exciting the acceptor. The acceptor then relaxes and may fluoresce. The donor yield becomesFor the acceptor, distinguish its intrinsic yield per acceptor excitation, $Q_A=k_{r,A}/(k_{r,A}+k_{nr,A})$, from the number of acceptor photons per donor excitation. Transfer increases the supply of acceptor excitations, giving sensitized emission $E{Q}_A$ per donor excitation in the ideal model. It does not by itself change $Q_A$ if the acceptor's own decay rates remain unchanged. Environmental changes, saturation or additional quenching can change those rates, but increased acceptor intensity alone does not establish an increased intrinsic fluorescence quantum yield.

Assume dilute nondegenerate electrons and electron holes, parabolic three-dimensional band edges with the stated scalar band effective masses, spin multiplicity two, one effective valley for each edge, and thermal equilibrium. Take the unchanged host ground state to have unit activity and neglect bound excitons. The effective density of states values areFor a Fermi level $\mu$ between the band edges,The factors of $\mu$ cancel on multiplication. With $E_g=E_c-E_v$, the semiconductor mass-action law isThis density-form equilibrium constant has units ${\mathrm{m}}^{-6}$. A dimensionless thermodynamic constant uses a specified reference number density $n^{\circ }$, giving $K=K_n/(n^{\circ }{)}^2$. The dot representing the host ground state is not a freely translating gas species requiring a further gas partition function. For an intrinsic semiconductor, charge neutrality gives $n_e=n_h=n_i=\sqrt{K_n}$.

Both in-plane orientations minimize the one-site energy. Since every adsorption site is occupied, the sites label independent choices even though the molecules are chemically identical. The ground state of the whole system therefore hasThe ground-state degeneracy is extensive: independently choosing either in-plane orientation at each site costs no energy. There is no extra vacancy or free-translational contribution in this model.

Only the two lower orientations remain populated, giving $2^N$ equiprobable microstates. ThusThis residual entropy follows from the stipulated independent-site ground-state degeneracy. The zero-entropy version of the third law of thermodynamics presupposes a unique ordered ground state; that assumption does not hold here. Intermolecular interactions or a small splitting of the two in-plane states could remove this degeneracy, but they are excluded from the given model.

The two protons are identical spin-one-half fermions, so the total state is antisymmetric under exchange. In the electronic and vibrational ground state, an even rotational $J$ has a symmetric spatial wavefunction and therefore uses the antisymmetric nuclear-spin singlet state; odd $J$ uses the symmetric triplet state. Thus parahydrogen has one nuclear-spin state, and orthohydrogen has three. Each rotational level also has spatial state degeneracy $2J+1$.

The translational partition functions and all common electronic/vibrational factors cancel in the ratio of equilibrium populations. Using the same zero of energy for the two spin species givesSubtracting the $J=1$ energy separately from all odd levels without restoring that offset in the equilibrium ratio would incorrectly remove the low-temperature preference for parahydrogen. The nuclear spin statistics account for the factor three, independently of the spatial factor $2J+1$.

The transition strength depends on $|R_{ji}{|}^2$, so a necessary condition is that at least one Cartesian electric dipole transition moment is nonzero. An integral can be nonzero only if its integrand contains the totally symmetric representation:The harmonic vibrational ground state is totally symmetric, and a one-quantum state of normal mode $k$ transforms as ${\Gamma }^{(k)}$. For the real irreducible representations here, the fundamental is therefore symmetry-allowed precisely whenEquivalently, expand the electric dipole moment near equilibrium:The fundamental has $⟨1_k|Q_k|0_k⟩\ne 0$, so vibrational infrared activity requires a nonzero dipole derivative of the matching symmetry. In $D_{6h}$, $z\sim A_{2u}$ and $(x,y)\sim E_{1u}$, giving $A_{2u}$ and $E_{1u}$ as the IR-active species. Symmetry permission does not guarantee a large band intensity.

The overall vibrational ground state is invariant under every molecular symmetry operation, so it transforms as $A_{1g}$. Combining it with the electronic ground state givesFor a degenerate harmonic normal mode, the zero-quantum wavefunction depends on the invariant sum $Q_x^2+Q_y^2$, so the ground vibrational symmetry remains totally symmetric even when excited vibrational states form multiplets.

A gives the same optimum: replacing $\alpha$ by $2\alpha$ merely reparameterizes the same set of positive-width Gaussian functions, so its best bound is unchanged.

B is worse as a ground-energy estimate: it has odd parity and a node at the origin, whereas the ground state of this even confining one-dimensional potential is even and nodeless. It is therefore orthogonal to the exact ground state and belongs to the odd sector, whose lowest energy is already above the ground energy; the extra factor $x$ cannot be removed by varying its width. A separate numerical minimization gives a best value about $1.92372$ for B, well above the original Gaussian function bound.

C can improve the bound: setting $\beta =0$ recovers every original Gaussian function, while nonzero $\beta$ adds shape freedom. Minimizing over this larger family can never give a worse optimum, and the additional freedom does lower the quartic-oscillator estimate. Inclusion of a family alone proves “no worse”; improvement is not guaranteed for every conceivable potential, for example one already solved exactly by a Gaussian function.

Within the well, the potential is zero and the Hamiltonian is purely kinetic:The infinite walls are encoded in the operator's boundary conditions $\Psi (0)=\Psi (a)=0$, rather than differentiated as part of the interior expression. Sincewe obtainThese positive energy eigenvalues grow quadratically with the quantum number; the ground state has a nonzero confinement energy.

The density $|{\Psi }_0{|}^2=\sqrt{\alpha /\pi }{e}^{-\alpha x^2}$ is even. Hence $⟨x⟩=0$; also ${\Psi }_0^{\prime }=-\alpha x{\Psi }_0$ gives $⟨p_x⟩=i\hslash \alpha ⟨x⟩=0$.

To evaluate the second position moment, differentiate the Gaussian integral with respect to its positive parameter:For the momentum second moment, apply the momentum operator twice:Both means vanish, soThe ground state attains the bound proved above and is therefore a minimum uncertainty state. A wider Gaussian reduces momentum spread while increasing position spread: the product remains at this lower limit. Other states, including excited oscillator eigenstates, generally give a larger product.